SIMPLIFY EXPRESSIONS INVOLVING EXPONENTIAL FUNCTION

Some more examples.

Simplify :

Example 1 :

(4^m + 8^m)/4^m

Solution :

=  (4^m + (4⋅2)^m)/4^m

=  (4^m + 4^m⋅2)^m)/4^m

=  [4^m(1 + 2^m)]/4^m

=  1 + 2^m

So, the answer is 1 + 2^m

Example 2 :

(4^m + 8^m)/(1 + 2^m)

Solution :

=  4^m+(4⋅2)^m/(1 + 2^m)

=  (4^m+4^m⋅2^m)/(1 + 2^m)

=  4^m(1 + 2^m)/(1 + 2^m)

=  4^m

So, the answer is 4^m

Example 3 :

(7^b + 21^b)/7^b

Solution :

Given, (7^b + 21^b)/7^b

=  (7^b + (7⋅3)^b)/7^b

=  (7^b + 7^b⋅3^b)/7^b

Factoring 7^b from the numerator, we get

=  [7^b(1 + 3^b)]/7^b

=  1 + 3^b

So, the answer is 1 + 3^b

Example 4 :

(4ⁿ⁺² – 4ⁿ)/4ⁿ

Solution :

Given, (4ⁿ⁺² – 4ⁿ)/4ⁿ

By using exponent of the product rule, we get

=  [(4ⁿ. 4²) – 4ⁿ]/4ⁿ

=  [(4ⁿ . 16) – 4ⁿ]/4ⁿ

=  [4ⁿ(16 – 1)]/4ⁿ

=  15

So, the answer is 15

Example 5 :

(4ⁿ⁺² – 4ⁿ)/15

Solution :

Given, (4ⁿ⁺² – 4ⁿ)/15

By using exponent of the product rule, we get

=  [(4ⁿ. 4²) – 4ⁿ]/15

Factoring 4ⁿ from the numerator, we get

=  [(4ⁿ . 16) – 4ⁿ]/15

=  [4ⁿ(16 – 1)]/15

=  4ⁿ(15)/15

=  4ⁿ

So, the answer is 4ⁿ

Example 6 :

(2^m+n – 2ⁿ)/2ⁿ

Solution :

Given, (2^m+n – 2ⁿ)/2ⁿ

By using exponent of the product rule, we get

=  [(2^m. 2ⁿ) – 2ⁿ]/2ⁿ

Factoring 2ⁿ from the numerator, we get

=  [2ⁿ(2^m – 1)]/2ⁿ

=  2^m - 1

So, the answer is 2^m – 1

Example 7 :

(3ⁿ⁺² – 3ⁿ)/3ⁿ⁺¹

Solution :

Given, (3ⁿ⁺² – 3ⁿ)/3ⁿ⁺¹

By using exponent of the product rule, we get

=  [(3ⁿ. 3²) – 3ⁿ]/(3ⁿ . 3¹)

=  [(3ⁿ . 9) – 3ⁿ]/(3ⁿ . 3)

=  [3ⁿ(9 – 1)]/[3ⁿ(3)]

=  8/3

=  2 2/3

So, the answer is 2 2/3

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